${x^{x\cos x}} + \cfrac{{{x^2} + 1}}{{{x^2} - 1}}$

Let$y = {x^{x\cos x}} + \cfrac{{{x^2} + 1}}{{{x^2} - 1}} = u + v,$

where $u = {x^{x\cos x}}$ and $v = \cfrac{{{x^2} + 1}}{{{x^2} - 1}}$
$\Rightarrow$ $\cfrac{{dy}}{{dx}} = \cfrac{{du}}{{dx}} + \cfrac{{dv}}{{dx}}$ …(i)

Now, $u = {x^{x\cos x}}$

By taking log on both sides , we get

$\log u = x\cos x \cdot \log x$ …(ii)

Differentiating (ii) w.r.t.x, we,get

$\cfrac{1}{u}\cfrac{{du}}{{dx}} = x\cos x\cfrac{d}{{dx}}(\log x) + x\log x\cfrac{d}{{dx}}\cos x + \cos x\log x\cfrac{d}{{dx}}(x)$

$\Rightarrow$ $\cfrac{1}{u}\cfrac{{du}}{{dx}} = x\cos x \cdot \cfrac{1}{x} + x\log x( - \sin x) + \cos x\log x$

$= \cos - x\sin x\log x + \cos x\log x$
therefore, $\cfrac{{du}}{{dx}} = {x^{x\cos x}}[\cos x - x\sin x\log x + \cos x\log x]$ …..(iii)

Also, $v = \cfrac{{{x^2} + 1}}{{{x^2} - 1}}$ ….(iv)

Differentiating (iv) w.r.t. x, we get

$\Rightarrow$ $\cfrac{{dv}}{{dx}} = \cfrac{{({x^2} - 1)\cfrac{d}{{dx}}({x^2} + 1) - ({x^2} + 1)\cfrac{d}{{dx}}({x^2} - 1)}}{{{{({x^2} - 1)}^2}}}$

$= \cfrac{{({x^2} - 1)(2x) - ({x^2} + 1)(2x)}}{{{{({x^2} - 1)}^2}}}$
$= \cfrac{{2x[{x^2} - 1 - {x^2} - 1]}}{{{{({x^2} - 1)}^2}}} = \cfrac{{ - 4x}}{{{{({x^2} - 1)}^2}}}$ …(v)

Substituting the values of (iii) \& (v) in (i), we get

$\cfrac{{dy}}{{dx}} = {x^{x\cos x}}[\cos x - x\sin x\log x + \cos x\log x] - \cfrac{{4x}}{{{{({x^2} - 1)}^2}}}$

$= {x^{x\cos x}}[\cos x(1 + \log x) - x\sin x\log x] - \cfrac{{4x}}{{{{({x^2} - 1)}^2}}}$